# Roll three dice, exactly two are even. What is the probability that exactly two are equal?

You roll three six-sided dice.

Exactly two of those three dice roll to even numbers.

What are the chances that exactly two of the dice rolled to the same number?

I have an answer, but I can't get it to agree with Bayes Theorem so I would especially appreciate help seeing how to use Bayes here.

Update: After receiving some unfriendly help below, I just want to clarify that my question was only about how one might use Bayes to think about this problem. While I understand that this is strictly unnecessary for solving the problem, I thought it might be interesting, and then I got intrigued when I had trouble using Bayes to agree with the simpler argument. That's the context for this question.

• What is the answer? Please provide as much information as possible. May 3 '19 at 17:31
• I think the answer should be 1/3. If you have two even dice, it doesn't matter what the first one is, and there are three things the second can be.
– MBP
May 3 '19 at 17:34
• If you are curious about Bayes, and you cannot get Bayes to give you 1/3, you should show us what you got with Bayes so we can debug it for you. May 3 '19 at 19:27
• The response that showed a correct way to use Bayes was very helpful and answered my question. I had miscalculated something -- which I knew was the problem, but I couldn't figure out what. Now I know -- thanks!
– MBP
May 3 '19 at 20:52

Yes, I am also getting $$\frac{1}{3}$$ as the answer. I have used Bayes' theorem because you explicitly asked for it. However, as you can see, the calculations can get quite messy and there are simpler ways to crack this problem (as other users have suggested).

Define the events $$A$$ and $$B$$ as follows:

$$A:$$ Exactly $$2$$ of the $$3$$ dice have the same number

$$B:$$ Exactly $$2$$ of the $$3$$ dice roll to even number \begin{align} P(A|B) &=\frac{P(B|A)\times P(A)}{\bigg(P(B|A)\times P(A)\bigg)+\bigg(P(B|A^c)\times P(A^c)\bigg)}\\ &=\frac{\frac{9}{30}\times \frac{{3\choose 2}\cdot6\cdot5}{6\cdot6\cdot6}}{\bigg(\frac{9}{30}\times \frac{{3\choose 2}\cdot6\cdot5}{6\cdot6\cdot6}\bigg)+\bigg(\frac{{3\choose2}\cdot3\cdot2\cdot3}{6\cdot6\cdot6-90 }\times\bigg(1-\frac{{3\choose2}\cdot6\cdot5}{6\cdot6\cdot6}\bigg)\bigg)}\\ &=\frac1{3} \end{align}

There is no need for Bayes here. Your sampling space once even numbers are observed is $$\{(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)\}$$. Out of $$9$$ possibilities $$3$$ are what you are after. Thus, the probability is $$1/3$$.

• Right, but I'm curious about Bayes.
– MBP
May 3 '19 at 18:56
• +1. and @MBP - if you are curious about Bayes, and you cannot get Bayes to give you $1/3$, you should show us what you got with Bayes so we can debug it for you. May 3 '19 at 19:26
• Sheesh, this community is not a terrific place to actually ask the questions you're interested in. I wanted to know about Bayes, I made that clear, I got an answer saying "you don't need Bayes," that wasn't an answer to my question, so I gently pressed the "-1." Also, it's a "-1" not the end of the world. You can handle it.
– MBP
May 3 '19 at 20:50
• I find this remarkable behavior on a Q&A site, and it definitely makes me less likely to post questions here in the future.
– MBP
May 3 '19 at 20:55
• @MBP: You downvoted gently? That's hilarious! Did you do it with love in your heart, out of a reluctant sense of duty? May 3 '19 at 21:07

Let $$I$$: $$2$$ identical numbers out of $$3$$; $$E$$: $$2$$ even numbers out of $$3$$. Then: $$P(I|E)=\frac{P(I\cap E)}{P(E)}=\frac{{3\choose 1}\cdot {3\choose 2}\cdot {3\choose 1}}{{3\choose 1}\cdot {3\choose 1}\cdot {3\choose 2}\cdot {3\choose 1}}=\frac13.$$ $$P(I\cap E)$$: Two identical even numbers and one odd number. There are $${3\choose 1}$$ ways to choose an even number, $${3\choose 2}$$ ways to allocate two such even numbers on three positions and $${3\choose 1}$$ ways to choose an odd number out of three odd numbers for the third position, hence $${3\choose 1}{3\choose 2}{3\choose 1}$$ ways to have two identical even numbers and one odd number.

$$P(E)$$: Two (any) even numbers and one odd number. There are $${3\choose 1}$$ ways to choose the first even number, $${3\choose 1}$$ ways to choose the second even number, $${3\choose 2}$$ ways to allocate the two even numbers on three positions and $${3\choose 1}$$ ways to choose an odd number for the third position, hence $${3\choose 1}\cdot {3\choose 1}\cdot {3\choose 2}\cdot {3\choose 1}$$ ways to have two even and one odd number.

• One thing this response is helping me realize is that I need to become more fluent in different versions of Bayes Theorem. Thank you!
– MBP
May 3 '19 at 21:13